Mystery #1

Slams with Minimal Trump Holdings

Question

This mystery is a group of related questions extending some proposed in The Bridge Journal in 1964. An answer to any of these questions should include a deal that illustrates the named holding.

Question 1A. If the East-West cards may be arranged as desired and play is double-dummy, what is the weakest trump holding with which North-South can make a grand slam?

Question 1B. Same as Question 1A, but East-West are denied a trump as the opening lead.

Question 1C. If the East-West cards may be arranged as desired and play is double-dummy, what is the weakest trump holding with which North-South can make a small slam?

Question 1D. Same as Question 1C, but East-West are denied a trump as the opening lead.

An answer to 1D was given by Terence Reese in Play Bridge With Reese. We conceal it here to allow you the pleasure of finding it for yourself if you are so inclined. Question 1C was posed, but not answered, in 1964.

Discussion

Comment from Jim Boyce:

The problems are not completely well-posed as stated. It may not be entirely clear whether one trump holding is weaker than another. One ordering of trump holdings that gives some interesting results is compare first the length and then highest and then second highest (and so on) of declarer's combined trump holding.

Reply from The Bridge World:

The results of Jim's "first attempts," as he puts them, will be found in the answer file. See the link, below. We are keeping the discussion and answer pages separate, so that no one will see an answer prematurely. Since this is a new arrangement, we are still in a state of flux regarding presentation, timing, and so forth. All comments and suggestions will be appreciated.

In Mystery # 1, we deliberately left the ambiguity in the problem, hoping someone would pick up the idea of whether the same holding satisfied more than one possible definition of "weakest." Note that if someone held 12 spades missing the king, that holding could be deemed weaker than any with the ace but including the king. (Also, this presentation preserved the original, ambiguous, form of the puzzle.)

The definition of "weakest" that seems to be most attractive to solvers is that shorter is weaker than longer, and within equal lengths a holding with a higher missing card is weaker than one with a corresponding lower missing card. A more formal approach to this definition is discussed within the solutions.

Solutions

The top part of the below solution summarizes the best solutions to Mystery #1 received through November, 1997. The astonishing final chapter, which occurred during December, appears at the end. Think of it as a test of will not to look at the ending until you have seen what preceded it.

(a) From Jim Boyce: Declarer can make seven spades on a combined trump holding of AKJ92. On the deal below, declarer, South, wins the first ten tricks, including one trump finesse (possibly on the opening lead) and nine side-suit winners, ending in dummy. He then leads any card.

	NORTH ♠ 2 ♥ A K Q 2 ♦ A K Q 2 ♣ A K Q 2
WEST ♠ 6 5 4 3 ♥ 8 7 6 ♦ 8 7 6 ♣ 8 7 6		EAST ♠ Q 10 8 7 ♥ J 10 9 ♦ J 10 9 ♣ J 10 9
	SOUTH ♠ A K J 9 ♥ 5 4 3 ♦ 5 4 3 ♣ 5 4 3

(b) No proposed solutions received through Thanksgiving--but see below.

(c) From Don Kersey: As Jim Boyce asserts, the problems are not well-posed. Looking for the moment at part (c), and assuming the question refers to a partnership's combined trump holdings only (so that, e.g., AJ9 opposite 82 is, for our purposes, the same as AJ98 opposite 2), then we can start with the following observation:

Suppose we impose a (weak) ordering on trump holdings by saying that holding A is strictly weaker than holding B if they are different and the following conditions are all true:

      --B has at least as many cards as A,
      --the highest card in A is at most equal to the highest card in B,
      --the second highest card in A is at most equal to the second highest card in B,
      --generally, the k-th highest card in A is at most equal to the k-th highest card in B, as k runs from 1 up to the length of A.

Then we can look for "minimal" slam-worthy trump holdings, that is, holdings that can lead to a made slam but for which no strictly weaker trump holding can lead to a made slam. I will show that there are exactly 9 such holdings:

Q J 10 9 8 7 6 5 4 3 2
K 9 8 7 6 5 4 3 2
K J 6 5 4 3 2
K J 9 4 3 2
K Q J 7 5
K Q 10 8 2
A J 9 8 2
A Q 9 3 2
A K J 7

Here is a proof that these nine holdings are minimal, and are the only minimal holdings--in other words, a trump holding can lead to a made slam if and only if it is one of these nine holdings or is strictly stronger than one of these nine holdings.

A general principle that makes part of our job easy is: If one of the defenders has more than n trumps, then declarer needs n of the 2n highest trumps in order to make slam. (So, for instance, if one defender has 3 trumps, then declarer needs 2 of the top 4 trumps).

This is quite easy to see: suppose declarer had fewer than n of the top 2n trumps; then the defenders would have at least n+1 of the top 2n trumps; since one of the defenders has at least n+1 cards in the suit, the defenders can always arrange to play their n+1 highest trumps on n+1 different tricks; declarer, with at most n-1 "high" trumps, can win at most n-1 of these tricks, leaving at least two tricks for the defenders.

An example may make this argument clearer: Suppose one of the defenders has at least 2 trumps; then if the defense has both the A and the K, they can (at double-dummy) always arrange not to crash their honors, and so they must be able to take at least two trump tricks.

Applying this principle to our problem, we see that any 11-card or longer holding can make slam, and any such holding is strictly stronger than (or equal to) Q J 10 9 8 7 6 5 4 3 2.

If declarer has 9 or 10 trumps, then the defenders have at least 3 trumps, and so at least one defender has at least 2 trumps, so (applying the principle) declarer must have 1 of the top 2 trumps to make a slam; the weakest conceivable holding is thus K 9 8 7 6 5 4 3 2, and we will see below that this can make slam. So, a 9 or 10 card trump holding can make slam if and only if it is strictly stronger than (or equal to) K 9 8 7 6 5 4 3 2.

If declarer has 7 or 8 trumps, then the defenders have at least 5 trumps, and so at least one defender has at least 2 trumps, so (applying the principle twice) declarer needs at least 1 of the top 2 trumps AND 2 of the top 4 trumps; the weakest conceivable holding is thus K J 6 5 4 3 2, and we will see that this does make slam. So a 7- or 8-card trump holding can make slam if and only if it is strictly stronger than (or equal to) K J 6 5 4 3 2.

If declarer has 6 trumps, then the defenders have at least 7, and, arguing as before, we see that declarer needs at least 1 of the top 2 AND at least 2 of the top 4 AND at least 3 of the top 6; the weakest conceivable holding is K J 9 4 3 2, and this can make slam. So a 6-card trump holding can make slam if and only if it is strictly stronger than (or equal to) K J 9 4 3 2.

Alas, this form of argument is inadequate to determine the answer for 5-card trump holdings, since K J 9 3 2 is not strong enough for slam, so special arguments are needed for trump holdings shorter than 6 cards. Since the case of 4 trumps is easier, let's look at that first.

If declarer has 4 (or fewer) trumps, then the defenders have at least 9; so, applying our principle 4 times shows that declarer needs 1 of the top 2 AND 2 of the top 4 AND 3 of the top 6 AND 4 of the top 8 (in particular, declarer needs 4 trumps, which is clear for other reasons as well). Note that this means declarer's smallest trump must be at least the 7. Obviously, neither defender can have more than 5 trumps, so the defensive trumps must split 5-4. If declarer's 4 trumps are split between North and South, then a trump lead by the defenders will leave declarer with only 2 trumps and the defenders with 7; declarer's two remaining trumps can account for at most 4 of the defenders' remaining trumps; this leaves the defenders with at least 3 trumps to play on tricks where declarer has no trump to play, so these 3 trumps must take at least 2 tricks. This shows that all four of declarer's trumps must be in the same hand, and trumps must be 5-4-4-0 around the table.

If the defenders have the ace of trumps, they can cash it at trick 1, and wait for a long trump trick (since one of the defenders has more trumps than declarer), so declarer must have the ace of trumps. If the defenders have either the K or both the Q and J, they can always take two tricks by following the strategy of never playing a trump until forced to do so (in effect, they make one trump trick by force of high cards, and one because of their extra length). Here, we show this in some detail for the defensive holding of K x x x x x x x x.

Suppose that declarer (with A Q J 10 of trumps, all in one hand) avoids leading a trump from hand in the early play (the argument is easier if declarer does play an early round of trumps). Then the defender with 5 trumps, by following the proposed strategy, will eventually be reduced to 5 trumps only. To have a chance, declarer mustn ot have expended any trumps to this point (if declarer has only 3 trumps and one defender has 5, then that defender gets 2 tricks automatically), and the other defender will also have an intact trump holding. At this point, declarer could lead from the void-trump hand and score one of the lower trump honors; but then declarer will be forced to play from the trump hand: if he exits with the last plain-suit card, this is ruffed low; declarer wins the trump return cheaply, but must then lead away from the AQ, conceding a second trick; nor can it help to play a trump first (a low trump would lose to the K, with a long trump to be lost later; cashing the ace before leading the side-suit card allows the defense to cash two tricks immediately). In effect, declarer can score one lower trump via a form of trump coup, and another via an endplay, but has no way to score the last lower trump. The argument with the defense's holding Q J x x x x x x x is similar.

Putting all the partial results everything together, we see that declarer needs both A and K, must have either the Q or J, and needs a lowest card at least the 7. The weakest holding meeting these conditions is A K J 7, and we will see that this holding does indeed make slam. Thus a 4-card trump holding can make slam if and only if it is strictly stronger than (or equal to) A K J 7.

It remains to consider 5-card trump holdings; this is the most technical part of the whole argument, and I give only a sketch. In this case it is enough to show two facts:

(i) all of A Q 9 3 2, A J 9 8 2, K Q 10 8 2, and K Q J 7 5 can make slam, while

(ii) none of K J 10 9 8, K Q 9 8 7, A J 10 7 6, K Q 10 7 6, K Q J 6 5, K Q J 7 4 can make slam.

To see that these two facts suffice, suppose declarer has a combined 5-card trump holding that can make slam, and remember that this holding must include at least 1 of the top 2, 2 of the top 4, and 3 of the top 6 trumps. Now:

(I) If the holding includes the A, then if the second card is the K or Q, the third card must be at least the 9, and the holding will be at least as strong as A Q 9 3 2, which is on list (i); otherwise, the second card must be the J, and the holding must also include either the 10 or the 9; since A J 10 7 6 cannot make slam, the fourth card must be at least the 8, so the holding will be at least as strong as A J 9 8 2, which is on list (i).

(II) If the holding doesn't include the A, then since neither K J 10 9 8 nor K Q 9 8 7 can make slam, the holding must include K Q, together with at least one of the J and 10. Since K Q J 6 5 can't make slam, the fourth card must be at least the 7; if the fourth card is the 8 or higher, then the holding is at least as strong as K Q 10 8 2, which is on list (i); so suppose the fourth card is the 7. Since K Q 10 7 6 cannot make slam, the third card must be the J; and since K Q J 7 4 cannot make slam, the fifth card must be at least the 5, and the holding is at least as strong as K Q J 7 5, which is on list (i).

It remains to show (i) and (ii). We look at (ii) first, and split it into two parts.

If the defenders have eight to the A Q or eight to the A J 10 between them, then they can always take two tricks by never playing a trump until forced to do so, which means that K J 10 9 8 and K Q 9 8 7 cannot make slam. The argument is similar (it involves no new ideas, but it splits into more cases) to the argument used above to show that nine to the king is always worth two tricks on defense.

For the other four holdings, the argument is a little tedious. We use the following result to eliminate all but a few special cases: Suppose that, after an opening trump lead, declarer is left with three trumps, or else with four trumps all in one hand, headed in either case by the 2nd and 3rd highest remaining trumps, while the defenders have six trumps including the 1st, 4th, and 5th highest remaining trumps. Then the defense can always take two trumps tricks, except in this special case: the 4th and 5th highest are doubleton in one hand, and declarer has the four-card holding including the 6th highest remaining trump.

To make the statement of this result and its proof more readable, pretend that the defenders' trumps are headed by A J 10, and declarer's by K Q, while the 6th highest remaining trump is the 9, the 7th highest the 8, and so on; we are saying that a defensive holding of A J 10 x x x can beat a holding of K Q x (x), unless the J 10 are doubleton and the K Q holding is at least K Q 9 x. (We could prove more, but this is all we need.)

To prove the result, suppose first that declarer makes slam by leading a round of trumps; this can't be a low trump, which would lose to the J or 10, with the A to be lost later, so it must be the K or Q; then, unless the J and 10 are doubleton, this trick will lose to the ace and will pick up only a small trump from the other defender, leaving the J and 10 as equals against declarer's remaining high card. Furthermore, even when the J 10 doubleton holding exists, declarer needs the 9 (by our usual principle: one of the defenders still has 4 trumps, so declarer needs 3 of the top 6); finally, unless declarer has a 4th trump, another trick will automatically be lost to the defender having A x x x. This gives us the exceptional case quoted in the result.

In any other case, declarer must avoid leading trumps. The defenders will also follow the policy of not playing trumps until forced to do so (after the initial trump lead). Suppose next that the remaining defensive trumps split 3-3. Then, since no one is playing trumps, the ending will eventually reduce to a 3-card ending in which both defenders have 3 trumps left (if declarer has the 4-card holding, then declarer will have scored one ruff by this point). Whether declarer now has 3 trumps in one hand, or 2 trumps in one hand and 1 in the other, the usual arguments show that the defenders can take two tricks by best play. Similarly, if the defensive trumps split 4-2 (but not J 10 opposite A x x x), consideration of cases shows that the defense can always take two tricks.

This result is almost good enough to eliminate the remaining 4 cases from (ii):

K Q J 6 5: the defense can lead trumps, playing the 7, 8, 9, or 10 to force an honor, leaving a winning position for the defense;

K Q J 7 4: the defense can lead trumps, playing the 8, 9, or 10 to produce a winning position for the defense unless all declarer's trumps are in one hand and one defender holds 10 9 8 exactly (then a special argument is required to show that the A 6 5 3 2 is still entitled to two tricks);

K Q 10 7 6: the defense can lead trumps, playing the 8, 9, or J to produce a winning position unless all declarer's trumps are in one hand and one defender holds J 9 8 exactly, in which case this trump lead would produce the exceptional case of the result;

A J 10 7 6: the defense can lead trumps, playing the Q or K to produce a winning position unless declarer ducks (then a special argument is required) OR one defender has K 9 8, Q 9 8, or 9 8 x exactly (then, again, any attempt to produce the position of the result will just lead to the exceptional case);

This leaves just a few explicit cases to be cleared up; the defense prevails in all of them by avoiding early trump plays.

We move now to the demonstration that the claimed minimal holdings can, indeed, produce slam.

It is easy to construct deals making slam for the 6-plus-card holdings. Essentially, we can always set up deals on which it is possible to draw trumps and claim. In the hardest case, give North 3 2, East A Q 10, and South K J 9 4; there, two leads from North will pick up the trump suit with only one loser). For the other cases, one pattern of suitable deals runs this way:

K Q J 7 5: All of these trumps must be in the same hand, say South, with West holding 10 9 8 and East A 6 4 3 2. Declarer can lead the K Q J, pinning the 10 9 8; South cashes 8 side winners finishing in North (defenders following throughout) and ends with a trump coup (7 5 over 6 4) if East has taken the A, or a coup en passant (7 5 over A 6) if East has ducked three times.

Example:

	NORTH ♠ — ♥ A K Q J ♦ A 5 4 3 ♣ A 5 4 3 2
WEST ♠ 10 9 8 ♥ 10 9 8 7 ♦ Q J 10 ♣ Q J 10		EAST ♠ A 6 4 3 2 ♥ 6 5 4 3 ♦ 9 8 ♣ 9 8
	SOUTH ♠ K Q J 7 5 ♥ 2 ♦ K 7 6 2 ♣ K 7 6

K Q 10 8 2: Give North the 2, East A J 9 x, and South K Q 10 8. Trumps can be led from North once; if East takes the A, then South cashes 9 side winners (defenders following) and ends with a trump coup; if East plays anything else, South wins cheaply and leads one of his remaining trump equals, then cashes his side winners and ends with a trump coup at trick 12 (if East took his A on the second round) or a coup en passant (if East ducked twice).

Example:

	NORTH ♠ 2 ♥ A K Q J ♦ A K 3 2 ♣ A 4 3 2
WEST ♠ 6 5 4 3 ♥ 10 9 8 7 ♦ J 10 9 ♣ Q J		EAST ♠ A J 9 7 ♥ 6 5 4 3 ♦ 8 7 6 ♣ 10 9
	SOUTH ♠ K Q 10 8 ♥ 2 ♦ Q 5 4 ♣ K 8 7 6 5

A J 9 8 2: Give North 8 2, East K Q 10, South A J 9. Trumps are led from North (or by the defenders on opening lead), and assuming East splits (it is easier otherwise), South wins and cashes 8 side winners, finishing in North. Then, South ruffs with the 9 while East follows suit and West underruffs; North ruffs with the 8 while East follows suit and West is overruffed; then, at trick 12, South scores the J en passant.

Example:

	NORTH ♠ 8 2 ♥ A K 4 3 2 ♦ A K 3 2 ♣ 3 2
WEST ♠ 7 6 5 4 3 ♥ J 10 9 ♦ J 10 9 ♣ Q J		EAST ♠ K Q 10 ♥ 8 7 6 ♦ 8 7 6 5 ♣ 10 9 8
	SOUTH ♠ A J 9 ♥ Q 5 ♦ Q 4 ♣ A K 7 6 5 4

A Q 9 3 2: Give North 3 2, East K J 10 x, South A Q 9. Trumps are led from North (or on opening lead), and we may as well assume East plays the 10, South's Q winning. Then, South cashes 8 side winners, ruffs a side-suit card with North's 3 (defenders following throughout), and, at trick 11, leads a plain suit from North. If East ruffs low, South scores the 9 immediately; if East ruffs high, South discards and wins two more trumps on the ensuing trump endplay.

Example:

	NORTH ♠ 3 2 ♥ A K 4 3 2 ♦ 2 ♣ A 5 4 3 2
WEST ♠ 7 6 5 4 ♥ J 10 9 ♦ J 10 9 8 ♣ Q J		EAST ♠ K J 10 8 ♥ 8 7 6 ♦ 7 6 5 4 ♣ 10 9
	SOUTH ♠ A Q 9 ♥ Q 5 ♦ A K Q 3 ♣ K 8 7 6

A K J 7: Give South all these cards, one opponent Q 10 9 8, the other 6 5 4 3 2. Assuming no trump lead (it is easier with a trump lead), South cashes 8 side winners (defenders following), ruffs something with the 7 while the long trump hand underruffs and the short trump hand follows suit, then exits with the last plain-suit card, which must be won by the Q 10 9 8 hand as it overruffs the other defender. Now, the Q 10 9 holding is endplayed in trumps.

Example:

	NORTH ♠ — ♥ A K 3 2 ♦ A K 4 3 2 ♣ A 4 3 2
WEST ♠ Q 10 9 8 ♥ J 10 9 8 ♦ J 10 9 ♣ Q J		EAST ♠ 6 5 4 3 2 ♥ 7 6 5 ♦ 8 7 6 ♣ 10 9
	SOUTH ♠ A K J 7 ♥ Q 4 ♦ Q 5 ♣ K 8 7 6 5

(d) [Editor's note: You may find it instructive to compare the contribution below with Reese's original solution.]

From Jim Boyce: If an opening trump lead is barred, declarer can make six spades on a combined trump holding of AQ107. On the deal below, declarer, South, cashes four hearts, two diamonds and two clubs, pitching two clubs and a diamond from dummy. Then, a club ruff with the seven, a diamond ruff cheaply, and a club exit.

	NORTH ♠ 7 ♥ 2 ♦ 8 7 6 5 4 3 2 ♣ 7 6 5 4
WEST ♠ 6 5 4 3 2 ♥ 6 5 4 3 ♦ 10 9 ♣ 9 8		EAST ♠ K J 9 8 ♥ 10 9 8 7 ♦ Q J ♣ Q J 10
	SOUTH ♠ A Q 10 ♥ A K Q J ♦ A K ♣ A K 3 2

The Final Demystification: During December, 1997, Don Kersey announced that he had solved the entire problem of weakest possible trump holdings, with or without a trump lead, not only for slam-level contracts but for all possible contracts. His complete solution appears on his web site.

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